Step 5: Solve the equation for g(x) by dividing by 2. 2g(x) 2 (x 4)2 2 4 2 g(x) 1 2 (x 4)2 2 From g(x) 1 2 (x 4)2 2 we can see that i) The vertex is (4, 2). ii) The axis of symmetry is x 4. iii) a 1 2 (the same as in the form g(x) 1 2 x2 4x 6) so the parabola opens downward. iv) Since a 1 2 , the graph of g(x) will be wider than y x2. Find some other points on the parabola. Use the axis of symmetry. y 1 2 2 x g(x) 6 0 8 6 Using the axis of symmetry, we can see that the x-intercepts are (6, 0) and (2, 0) and that the y-intercept is (0, 6). The domain is (q, q); the range is (q, 2. x YOU TRY 5 Graph each function. Begin by completing the square to rewrite each function in the form f (x) a(x h)2 k. Include the intercepts. a) f (x) x2 4x 3 b) g(x) 2x2 12x 8 4 Graph f(x) ax2 bx c Using a b 2a , f a b 2a bb We can also graph quadratic functions of the form f (x) ax2 bx c by using the formula h b 2a to fi nd the x-coordinate of the vertex. This formula comes from completing the square on f (x) ax2 bx c. Although there is a formula for k, it is only necessary to remember the formula for h. The y-coordinate of the vertex, then, is k f a b 2a b. The axis of symmetry is x h. Property The Vertex Formula The vertex of the graph of f (x) ax2 bx c (a 0) has coordinates a b 2a , f a b 2a bb. www.mhhe.com/messersmith SECTION 10.5 Quadratic Functions and Their Graphs 661
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